How do you graph `f(x) = (x-2)^2 - 1` on a coordinate graph?

Graph:

`f(x)=(x-2)^2-1` is a quadratic equation in vertex form:

`f(x)=a(x-h)^2+k`,

where:

`a=0`, `h=2`, `k=-1`

In order to graph a parabola, the vertex, x- and y-intercepts, and additional points need to be determined.

Vertex: maximum or minimum point of a parabola

The vertex is the point `(h,k)`: `(2,-1)`

Y-intercept: value of `y` when `x=0`

Substitute `y` for `f(x)` and substitute `0` for `x`. Solve for `y`.

`y=(0-2)^2-1`

`y=(-2)^2-1`

`y=4-1`

`y=3`

The y-intercept is `(0,3)`.

X-intercept: values of `x` when `y=0`

Substitute `0` for `y` and solve for `x`.

`0=(x-2)^2-1`

Expand.

`0=x^2-4x+4-1`

Simplify.

`0=x^2-4x+3`

Factor.

`0=(x-1)(x-3)`

Set each binomial equal to `0` and solve for `x`.

`x-1=0`

`x=1`

`x-3=0`

`x=3`

x-intercepts (roots): `(1,0)` and `(3,0)`

Additional point: choose value of `x`, plug it into the equation and solve for `y`.

`x=4`

Substitute `4` for `x` and solve for `y`.

`y=(4-2)^2-1`

`y=2^2-1`

`y=4-1`

`y=3`

Point: `(4,3)`

Summary

Vertex: `(2,-1)`

Y-intercept: `(0,3)`

X-intercepts: `(1,0)`, `(3,0)`

Additional point: `(4,3)`

Plot the points and sketch a parabola through them. Do not connect the dots.

graph{y=(x-2)^2-1 [-10, 10, -5, 5]}